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1 year ago

What causes a decrease in an objects weight

Physics

1 answer:

Artist 52 [7]1 year ago

4 0

The mass of an object can neither be change but the weight of the object can be changed.

<h3>What causes a decrease in an object weight?</h3>

The weight of the body or of an object can be changed if the body is placed farther away from the earth or placed in the planet which is far away from the earth's gravitational field, so the force of gravity on the object will change. However the mass of the body or mass of the object will remains the same regardless of whether the object is on Earth, in outer space, or on the Moon. By doing so the weight of the object or body will change but the mass remains the same.

So we can conclude that: The mass of an object can neither be change but the weight of the object can be changed.

Learn more about Weight here: brainly.com/question/86444

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Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with

Citrus2011 [14]

Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

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towards the sister (-F) (greater force applied)

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3 years ago

A pitcher claims he can throw a 0.146-kg baseball with as much momentum as a 2.70-g bullet moving with a speed of 1.50 ✕ 103 m/s

Nataly_w [17]

(a) The pitcher must throw the ball at 27.7 m/s

The momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is the object's velocity

Let's calculate the momentum of the bullet, which has a mass of

m = 2.70 g = 0.0027 kg

and a velocity of

$v=1.50\cdot 10^3 m/s$

Its momentum is:

$p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s$

The pitcher must throw the baseball with this same momentum. The mass of the ball is

m = 0.146 kg

So the velocity of the ball must be

$v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s$

So, the pitcher must throw the ball at 27.7 m/s.

(b) a. The bullet has greater kinetic energy

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where m is the mass of the object and v is its speed.

For the bullet, we have:

$K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J$

For the ball:

$K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J$

So, the bullet has greater kinetic energy.

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3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?